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66 Chapter 2. On a conjecture about addition modulo 2 k − 1equality of the remaining terms, or that ∆ defined as⎛∆ = ∑ (S + k n )!⎝ ∑ 2 k [ ] ⎞ ⎛h − k ⎠ C|k n − 1|! (h − k)! S + k kn−1 − (S + 1)! ⎝ ∑nk n≥0k≥1k≥1⎛− ∑ (S + k n + 1)!⎝ ∑ 2 k [ ] ⎞ h − k ⎠ (A|k n − 1|! (h − k)! S + k n + 1kn − 3 kn=0)k≥1k n≥0is zero. We split out the first two terms of the first sum on k n :⎛2 k(h − k)!S! ⎝ ∑ k≥1[ ] ⎞ ⎛h − k ⎠ − 13 S 6 (S + 1)! ⎝ ∑ k≥12 k(h − k)!2 k(h + 1 − k)![ ] ⎞ h − k ⎠ ,S + 1[ ] ⎞ h + 1 − k ⎠S + 1and the first one of the second sum on k n :⎛(S + 1)! ⎝ ∑ k≥12 k(h − k)![ ] ⎞ ( )h − k ⎠ 1S + 1 3 − 3,so that ∆ becomes∆ = ∑k n≥2⎛(S + k n )!⎝ ∑|k n − 1|!k≥1⎛2 k+ S! ⎝ ∑ (h − k)!k≥1⎛− (S + 1)! ⎝ ∑ (h + 1 − k)!k≥1⎛− ∑ (S + k n + 1)!⎝ ∑|k n − 1|!k n≥1k≥12 k [ ] ⎞ ( h − k ⎠ A(h − k)! S + k kn−1 + B )k nn k n[ ] ⎞ ⎛h − k ⎠ + 1 S 2 (S + 1)! ⎝ ∑ k≥12 k[ ] ⎞h + 1 − k ⎠S + 12 k [(h − k)!h − kS + k n + 12 k(h − k)!] ⎞ ⎠ A kn .[ ] ⎞ h − k ⎠S + 1Making the change of summation variable k n = k n + 1 in the second sum on k n , the terms inA kn cancel out between the two sums on k n and we get(∑ (S + k n)! ∑[ ] ) (2 k h − k∑[ ] )2 k h − k∆ =Bk n! (h − k)! S + kkn + B 0S!n (h − k)! Sk n≥2k≥1k≥1( ∑[ ] ) (2 k h − k∑[ ] )2 k h + 1 − k+ B 1(S + 1)!− (S + 1)!(h − k)! S + 1(h + 1 − k)! S + 1k≥1k≥1(∑ (S + k n)! ∑[ ] ) (2 k h − k∑[ ] )2 k h + 1 − k=Bk n! (h − k)! S + kkn − (S + 1)!n (h + 1 − k)! S + 1k n≥0k≥1k≥1(∑ 2 k ∑( ) [ ] ) (S + kn h − k∑[ ] )2 k h + 1 − k= S!B kn − (S + 1)!(h − k)!S S + k n (h + 1 − k)! S + 1k≥1k n≥0k≥1(∑ 2 k ∑( ) [ ][ ] )S + kn h − k= S!B kn − S + 1 h + 1 − k.(h − k)!S S + k n h + 1 − k S + 1k≥1k n≥0